$A (6, 1), B (8, 2)$ and $C (9, 4)$ are three vertices of a parallelogram $ABCD$. If $E$ is the mid-point of $DC$, find the area of $\triangle ADE$.

Given:

$A (6, 1), B (8, 2)$ and $C (9, 4)$ are three vertices of a parallelogram $ABCD$.

$E$ is the mid-point of $DC$.

To do:

We have to find the area of $\triangle ADE$.

Solution:

Let the fourth vertex of the parallelogram be (x, y).

We know that, the diagonal of a parallelogram bisect each other.

The mid-point of a line segment joining the points \( (x_{1}, y_{1}) \) and \( (x_{2}, y_{2})=(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}) \)

Mid-point of \( \mathrm{BD}= \) Mid point of \( \mathrm{AC} \)

\( \Rightarrow (\frac{8+x}{2}, \frac{2+y}{2}) \)

\( =(\frac{6+9}{2}, \frac{1+4}{2}) \)

\( \Rightarrow (\frac{8+x}{2}, \frac{2+y}{2})=(\frac{15}{2}, \frac{5}{2}) \)

On comparing, we get,

\( \frac{8+x}{2}=\frac{15}{2} \)

\( \Rightarrow 8+x=15 \)

\( \Rightarrow x=15-8=7 \)

\( \frac{2+y}{2}=\frac{5}{2} \)

\( \Rightarrow 2+y=5 \)

\( \Rightarrow y=5-2=3 \)

The fourth vertex of the parallelogram is \( \mathrm{D}(7,3) \).

The mid-point of side \( \mathrm{DC}=(\frac{7+9}{2}, \frac{3+4}{2}) \)

\( E=(8, \frac{7}{2}) \)

The area of \( \Delta \mathrm{ABC} \) with vertices $(x_{1}, y_{1}),(x_{2}, y_{2})$ and $(x_{3}, y_{3})=\frac{1}{2}[x_{1}(y_2-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$.

This implies,

Area of \( \Delta \mathrm{ADE} \) with vertices \( \mathrm{A}(6,1), \mathrm{D}(7,3) \) and \( \mathrm{E} (8, \frac{7}{2}) \)

\( =\frac{1}{2}[6(3-{7}{2})+7(\frac{7}{2}-1)+8(1-3)] \)

\( =\frac{1}{2}[6 \times(\frac{-1}{2})+7(\frac{5}{2})+8(-2)] \)

\( =\frac{1}{2}(-3+\frac{35}{2}-16) \)

\( =\frac{1}{2}(\frac{35}{2}-19) \)

\( =\frac{1}{2}(\frac{-3}{2}) \)

\( =\frac{-3}{4} \)

Area cannot be negative.

Hence, the required area of \( \Delta \mathrm{ADE} \) is \( \frac{3}{4} \) sq. units.